2thatistheaverageofaandnaWithn3anda2wewanttheaverageof2and32whichis1and34thatis145Nowwithnstill3butthebetterapproximation145toitssquarerootapplythemethodagaintogetastillbetterapprox ID: 607924
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Thereciprocalof18is0;03,20,andtheproductof50and0;03,20is2;46,40.Thereciprocalof32is0;01,52,30,andtheproductof0;01,22,30and1,21is2;31,52,30.Exercise24.ConverttheBabylonianapproximation1;24,51,10ofthesquarerootof2todecimalsanddeterminetheaccuracyoftheapprocimation.Thisismosteasilycomputedfromrighttoleft:((10=60+51)=60+24)=60+1=1:414212963whichisprettycloseto1.414213562.Sixdigitsofaccuracy,andonlyo by1inthenextdigit.Exercise25.UsetheassumedBabyloniansquarerootal-gorithmofthetexttoshowthatthesquarerootof3isabout1;45bybeginningwiththevalue2.Findathree-sexagesimal-placeapproximationtothereciprocalof1;45anduseittocalculateathree-sexagesimal-placeapproxima-tiontothesquarerootof3.Themethodsaysstartingwithanapproximaterootaofn,abetterapproximationwouldbea+b=(2a)whereb=n�a2.Ifweusealittlemodernalgebra,wecansimplifytheexpressionforthebetterapproximationtosimplya+n=a 2,thatis,theaverageofaandn=a.Withn=3,anda=2,wewanttheaverageof2and3=2,whichis1and3=4,thatis,1;45.Now,withnstill3,butthebetterapproximation1;45toitssquareroot,applythemethodagaintogetastillbetterapproximation.Wewanttheaverageof1;45and3multipliedbythereciprocalof1;45.Now1;45is7/4,sowewant4/7insexagesimal.Therearevariouswaystondthatinsexagesimal.Ifyou'rewillingtouseamoderncalculator,punchin4/7.Multiplyby60andyouseeyouget34plusafraction.Subtractthe34andmultiplyby60again.Youget17plusafraction.Subtractthe17andmultiplyby60again.Yougetjustunder9.Thus,thereciprocalisabout;34,17,09.Multiplythatbyn=3toget1;42,51,27.Nowaveragethatwith1;45.Addthetwotogethertoget3;27,51,27,thenhalvetoget1;43,55,43,30.Decimally,that's1.73214.Asp 3=1:732050808to10sig-nicantdigits,thisancientBabylonianapproximationwasaccurateto4signicantdigits.Incidentally,thisalgorithmworksverywell,eachiterationdoublesthenumberofdigitsofaccuracy.OnemoreiterationwouldhavegiventheBabylonians8digitsofaccuracy.Exercise28.SolvetheproblemfromtheOldBabyloniantabletBM13901:Thesumoftheareasoftwosquaresis1525.Thesideofthesecondsquareis2=3thatoftherstplus5.ndthesidesofeachsquare.(Youmayusemodernmethodstondthesolution.)Ifthesidesofthetwosquaresarexandy,thenwehavetwoequationsx2+y2=1525y=2 3x+5Eliminatingygivesoneequationinxthatsimpliesto13 9x2+20 3x�1500=0.Youcansolvethisbystandardmethodstogetx=30.(Thenegativesolutionisnotfeasi-ble.)Sothetwosquareshavesides30and25.Math105HomePageathttp://math.clarku.edu/~djoyce/ma105/2